Control Structures. Flowchart Symbols

Transcription

1 Control Structures The basic operations of a C/C++ programming language are specified using the following statements or structures: the input statement the output statement the assignment statements the single-way selection structure the two-way selection structure the multiple-way selection structure the counter-controlled iteration structure, and the logically-controlled iteration structure. The single-way selection, the two-way selection, the multiple-way selection, the counter-controlled iteration and the logically-controlled iteration are collectively called control structures. The specification of a controlled structure depends on a logical expression called condition. Controlled structures may also be specified using the following flowchart symbols: Flowchart Symbols SYMBOL NAME DESCRIPTION (a) Terminal Indicates the beginning or the end of an algorithm. (b) (c) Input/Output Process Indicates an input or an output operation. Indicates a computation or a data manipulation. (d) (e) Flow Lines Decision Used to connect the symbols and to indicate the logic flow. Indicates a decision point in an algorithm. (f) Connector Indicates an entry or an exit from another part of the flowchart Gilbert Ndjatou Page 57

2 Logical Expressions A logical expression (or condition) is an expression that evaluates to either true or false. There are two types of conditions: simple conditions and Compound conditions. Simple Conditions A simple condition has the following syntax: <arithmetic expression> <relational operator> <arithmetic expression> Relational Operators C/C++ Symbol Meaning < is less than > is greater than == is equal to <= is less than or equal to >= is greater than or equal to!= is not equal to Evaluations of simple conditions Assuming that the variables are defined and initialized as follows: int num1 = 5, num2 = 7, num3 = 2; float fnum = 11.75; char ch = K ; Evaluate the following conditions: a) num1 >= 5 d) num2 + 3 == num3 * 4 b) fnum + 7.2!= fnum / 2 e) 3 * num1 + 4 < num3 * 2 c) fnum * 2 > num f) A <= ch 2013 Gilbert Ndjatou Page 58

3 Solutions a) num1 >= 5 d) num2 + 3 == num3 * 4 5 >= == 2 * 4 10 == 8 b) fnum + 7.2!= fnum / 2 e) 3 * num1 + 4 < num3 * != / 2 3 * < 2 * != < 4 c) fnum * 2 > num1 +20 f) A <= ch *2 > A <= K > 25 Characters are ordered according to their ASCII code representations: 0' < 1' < 2' <... < 9' < A <... < Z < a... < z. Relational operators have a lower precedence than arithmetic operators: o In the absence of parentheses, arithmetic operations are evaluated before any relational operation is evaluated. However, it is a good programming practice to use parentheses to clarify the meaning of logical expressions. For example: (fnum * 2) > (num1 + 20) Compound Conditions A compound condition is built from simple conditions using logical operators. Logical Operators C++ Symbol Meaning Evaluation && AND OR <condition1> && <condition2> is true if and only if both conditions are true <condition1> <condition2> is true if and only if at least one of the two conditions is true! NOT!<condition> is true if and only if <condition> is false 2013 Gilbert Ndjatou Page 59

4 Evaluations of compound conditions && && && && There is a short-circuit evaluation of a compound condition when you do not have to evaluate the right-most condition in order to get the true value of the compound condition: and Therefore - Also, && and && Therefore && - Exercise CS1* Assuming that the variables are defined and initialized as follows: int num1 = 5, num2 = 7, num3 = 2; float fnum = 11.75; char ch = K ; Compute the true value of each of the following compound conditions: a) num1 >= 5 num2 + 3 == num3 d) num2 + 3 == num3 * 4 ch >= Z b) A <= ch && fnum + 7.2!= 2.5 e) num1 < 3 && num2 > num3 c)!(3 * num1 + 4 < num3 * 2) f)!(fnum * 2 < num1 + 20) 2013 Gilbert Ndjatou Page 60

5 Precedence of C/C++ Operators Operator Order of Evaluation Precedence! Unary right to left 7 * / left to right 6 % + - left to right 5 < <= > left to right 4 <= == left to right!= 3 && left to right 2 left to right 1 In the evaluation of an expression, operators with higher precedence are evaluated before those with lower precedence. Operators with the same precedence are evaluated in the order specified in the order of evaluation column. Exercise CS2 Assuming that the variables are defined and initialized as follows: int num1 = 9, num2 = 5, num3 = 10; float fnum = 12.50; char ch = P ; Evaluate the following conditions (using short circuit evaluation whenever possible): a) 2 * num1-5 >= 9 fnum / <= 6.5 b) num1 + num2 == num3 + 5 && 2 * num3 <= 4 * num2 c)! (num1 + 5 <= 13) d) num1 + 5 = = 24 num1 1 > num2 5 e) 2 * num1-5 >= 9 && fnum / <= 6.5 f) num1 - num2 <= num1 / 3 num1 * num2 > 100 g) num1-5 >= num3 num2 < 15 && num1 >= Gilbert Ndjatou Page 61

6 Accuracy of Floating-Point Values The fact that floating-point values are approximated inside a computer makes it difficult to test for the equality of floating-point values. For example, if the variable fnum is defined as follows: float fnum = 7.1; Then, the condition: fnum / 2 == 3.55 may not be true. The problem may be solved by assuming that two floating-point values are equal if their difference is relatively very small. This is done by testing if the absolute value of their difference is less than a certain value chosen by the programmer (for example ). Using the library function fabs() that takes as argument a floating-point value and returns its absolute value, the condition: value1 == value2 is replaced with the condition: fabs(value1 - value2) < which tests whether the difference between the two values is small enough so that we can make them equal. Evaluation of and In the C/C++ programming language, false is represented by the integer value 0, and any value other than 0 represents true. Therefore, the following two conditions are equivalent (have the same true value): o Conditions value!= 0 and value (value is not zero means that it is true) o Conditions value == 0 and! value (value is zero means that!value is true) bool Data Type A variable with data type bool can only hold the bool values true and false. Examples: bool flag = true; bool condvalue = false, HighTemperature, ExtremeTemperature; double temperature; HighTemperature = false cin >> temperature; HighTemperature = (temperature >= 120) ExtremeTemperature = (HighTemperature (temperature <= -20); 2013 Gilbert Ndjatou Page 62

7 Exercise CS3 1. Which of the following conditions are equivalent (that means have the same true value)? a) num!= 0 d)!num b) num == 0 e)!(num == 0) c) num 2. Suppose that a person s age is stored in the variable age, his number of children in the variable numchild, his salary in the variable salary, and his height in the variable height. Write the relational expressions to specify the following conditions: a. he is 45 year old. b. he is more than 5.7 feet tall. c. his salary is between 35,000 and 50,000. d. he does not have 3 children. e. he is either 6.0 feet tall, or he has less than 4 children. f. he is not older than 35 and he has 2 or 3 children 2013 Gilbert Ndjatou Page 63

8 Two-Way Selection using the if-else Structure The solutions of some problems require the CPU to test a condition and to select the action to be performed based on whether or not that condition is true. Example: Suppose that at a party, you serve alcohol drinks only to guests who are 21 or older, and a juice to guests who are less than 21. When a guest comes to get a drink, you first test his age: if he is 21 or older, you serve him an alcohol drink; otherwise you serve him a juice. The solutions of these types of problems are written using a two-way selection structure which is specified using a flowchart as follows: Condition F-Statement T-Statement Next-Statement It says to do the following: Test condition a. if it is true, perform the actions (T-Action) specified by T-Statement, b. otherwise, perform the actions (F-Action) specified by F-Statement. Next-Statement is the statement that says what to do after any of the actions (F-Action or T-Action) above has been executed. A two-way selection structure is specified in C++ using the if - else structure as follows: if (<condition>) <T-statement> else <F-statement> <Next-Statement> 2013 Gilbert Ndjatou Page 64

9 <T-statement> is a single statement that specifies the T-Action <F-statement> is a single statement that specifies the <F-Action>. Case Study CS1 Problem Statement Write a program to read the age of an individual, and to output the message serve alcohol drink if he is 21 or older, and the message serve juice otherwise. Your program will then output the message thank you for using this program. Program Logic output: Serve alcohol or Serve Juice, depending on the individual s age. Input: an individual s age. Variables: age (int) to hold an individual s age. Algorithm Specification (using a Flowchart) start Read age Write serve a juice age >= 21 Write serve alcohol drink Write Thank you for using this program Stop 2013 Gilbert Ndjatou Page 65

10 Figure CS1 Using the if-else Structure /************************************************************************* Program to read the age of an individual and to output the type of drink that he should be served *************************************************************************/ #include <iostream> using namespace std; #define DRINKAGE 21 int main() int age; // to hold an individual s age /* read the individual s age */ cout << \n\nenter the individual s age please:\t ; cin >> age; /* determine what type of drink he should be served */ if (age >= DRINKAGE) // he is over the drinking age cout << endl << Serve alcohol drink ; else // he can not drink alcohol cout << endl << Serve juice ; cout << endl << Thank you for using this program ; return (0); Case Study CS2 Problem Statement Write a program to read a positive integer value and to determine if it is even or odd. If it is even, print it with the message EVEN ; otherwise, print it with the message ODD. Program Logic Output: Input: Variable: the input value with the message EVEN or ODD, depending on whether the input value is even or odd. an integer value. num (int) to hold the input value. Note: An integer value is even if the remainder in its division by 2 is 0; otherwise, it is odd Gilbert Ndjatou Page 66

11 Algorithm Specification (using a Flowchart) Start Read num num % 2 = 0 Write num, odd Write num, even Stop Figure CS2 Using the if - else Structure /************************************************************************ Program to read an integer value and to determine if it is even or odd ************************************************************************/ #include <iostream> using namespace std; int main() int num; // to hold the value read /* read in an integer value */ cout << \n\nenter an integer value please:\t ; cin >> num; /* determine if it is even or odd */ if (num %2 == 0) // it is even cout << endl << num << \teven ; else // it is odd cout << endl << num << \todd ; return ( 0 ); Exercise CS4 Write a C++ code segment to read an integer value and to determine if it is a multiple of 5. If it is a multiple of 5, print it with the message MULTIPLE OF 5, otherwise, print it with the message NOT MULTIPLE OF Gilbert Ndjatou Page 67

12 Exercise CS5 Write a C++ code segment to read a character and to determine if it is a letter of the alphabet. If it is a letter of the alphabet, print it with the message, it is a letter otherwise, print it with the message, it is not a letter. Compound Statement A compound statement has the following syntax: (One or more statements to be executed as a block) Example num2 = num1 + 5; num1 = num1 +1; cout << endl << num1=\t << num1 << \n num2=\t << num2; The statements of a compound statement are executed one after another, starting with the first one in the sequence. A compound statement may appear anywhere in a C++ program where a single statement may appear. If you have to specify two or more statements as the T-statement or the F-statement of an if-else structure, these statements must be specified as a compound statement. Case study CS3 Problem Statement Write a program to read an integer value and to do the following: If the value read is zero, print it with the message INVALID DIVISOR. Otherwise read a second value and then compute and print the quotient and the remainder in the division of the second value by the first. At the end write the message Thank you for using this program Gilbert Ndjatou Page 68

13 Program Logic Output: Input: Variables: the quotient and the remainder in the division of the second value by the first or the message 0 is invalid divisor depending on the input. one or two integer values, depending on the value of the first. divisor (int) to hold the first value. dividend (int) to hold the second value. Note: The second value is input and the quotient and the remainder in the division of the second value by the first are computed only if the first value is not zero. Algorithm Specification (using a Flowchart) Start Read divisor divisor == 0 Read dividend Write dividend / divisor dividend % divisor Write divisor invalid divisor Write Thank you Stop 2013 Gilbert Ndjatou Page 69

14 Figure CS3 Using a Compound Statement in an if-else Structure /************************************************************************* Program to read two integer values and to compute the quotient and the remainder in the division of the second value by the first. *************************************************************************/ #include <iostream> using namespace std; int main() int divisor, dividend; // to hold the first value // to hold the second value /* read in the divisor */ cout << \n\nenter the divisor please:\t ; cin >> divisor; /*-----read the dividend and compute the quotient and the remainder--*/ if (divisor == 0) // it is invalid cout << endl << divisor << \tis INVALID DIVISOR ; else cout << \n\nenter the dividend please:\t ; cin >> dividend; cout << \nthe quotient in the division of:\t << dividend << by << divisor << \tis:\t << (dividend / divisor); cout << \nand the remainder is:\t << (dividend % divisor); cout << \n\nthank you for using this program ; return( 0 ); Exercise CS6* Assuming that all variables are properly defined and initialized, what are the error(s) in each of the following program segments? a. cin >> num; if (num = 5) sum = num + 4; else sum = num + 10; b. cin >> num; if (num > 5 && <= 10) cout << (num + 15); else cout << (num - 3); 2013 Gilbert Ndjatou Page 70

15 c. cin >> num; if (num < 15) result1 = 2 * num; result2 = num + 20; else result1 = 5 * num; result = result1 + result2; cout << \nthe result is:\t << result; Exercise CS7* Trace the execution of the following program segment and show its output for each of the following input values: a. input: 4 b. input: 20 Line # Statements 1 cin >> num; 2 if (num <10 ) 3 num = num + 6; 4 cout << endl << num =\t << num; else 5 cout << endl << num / 4 =\t << (num 4); 6 cout << endl << result =\t << (2 * num); Exercise CS8 Write a C++ code segment to read an integer value into variable num1and another integer value into variable num2 and to do the following: if the value in variable num1 is greater than the value in variable num2 do the following: add 1 to the value in num1 and 5 to the value in num2 and then print the result of the product of the new value in num1 and the new value in num2; otherwise, do the following: subtract 1from the value in num1 and 5 from the value in num2 and then print the result of the product of the new value in num1 and the new value in num2. Exercise CS9 A department store offers a rebate of 5% if the total amount of the purchase is at least 100 and a rebate of 2% otherwise. A sale tax of 8.25 % is also imposed on the price of each purchase (after the rebate). Write a C++ code segment to read the unit price and the number of items purchased, and to output the amount of the rebate and the price (including taxes) of the purchase Gilbert Ndjatou Page 71

16 Counter-Controlled Iteration using the while Structure The solutions of some problems require the CPU to repeat the execution of one or more statements a certain number of times. Example: Suppose that a candy bar costs 70 cents in a candy machine that only accepts dimes. To buy a candy bar from this machine, a user has to deposit a dime in the machine seven times. The solutions of these types of problems are written using a counter-controlled iteration (or loop) which is specified using a flowchart as follows: counter = initial-value Counter < final-value Body-of-the-Loop Next-Statement Body-of- the- Loop counter Next-Statement consists of the statements whose executions are repeated by the CPU. is called the loop counter it is a variable that is used to hold the number of repetitions is the first statement to be executed after the repetitions. Initial/Final Value If the initial value is 0 (no execution is done yet), then the final value must be the number of executions. If the initial value is 1 (one execution is about to be done), then the final value must be the number of execution plus 1. Loop Increment After each execution of the body-of-the-loop, the loop counter must be updated to reflect the number of executions. The statement that you use to update the loop counter is called the loop increment. The loop increment is a statement in the body-of-the-loop Gilbert Ndjatou Page 72

17 Algorithm of the Candy Bar Machine (using a flowchart) Start total = 0 count = 0 count < 7 Read coin total = total + coin Write Thank you count = count + 1 Stop Specification of a Counter-Controlled Iteration in C++ A counter-controlled iteration may be specified in C++ using the while structure as follows: counter = initial-value; while (counter < final-value) <Body-of-of the-loop (including the increment-statement)> <Next-statement> 2013 Gilbert Ndjatou Page 73

18 Algorithm for the Candy Bar Machine Problem (in C++) int total, coin, count; // to hold the total value of the coins // to hold the value of a coin // to hold the number of repetitions total = 0; count = 0; while ( count < 7 ) cin >> coin; total += coin; count ++; cout << endl << Thank You ; Case Study CS4 Problem Statement Write a program to read 30 weight measurements in pounds and to convert them into kilograms. Note that 1 Lb =.454 Kgs. Program Logic Input: Output: Variables: 30 weight measurements (in Lbs). 30 weight measurements (in Kgs). count (int) to count the weight measurements: Initial value is 0 - Final value is 30. pound (double) to hold a weight measurement in Lbs Gilbert Ndjatou Page 74

19 Algorithm Specification (using a flowchart) Start count = 0 count < 30 Read pound Write.454 * pound count = count + 1 Write Thank you Stop Figure CS4 Counter-Controlled Iteration using the while Structure Line Number 1 /************************************************************* 2 Program to read 30 weight measurements in pounds and convert 3 them into kilograms 4 *************************************************************/ 5 #include <iostream.h> 6 #include <iomanip.h> 7 #define MAXCOUNT 30 8 #define COEFICIENT int main() int count; // to count weight measurements 13 double pound, // the current weight measurement in Lbs Gilbert Ndjatou Page 75

20 15 cout << fixed << showpoint << setprecision(2); /*read all weight measurements (Lbs) and convert them to Kgs*/ 18 count = 0; // no weight measurement is read so far 19 while (count < MAXCOUNT) // repeat thirty times cout << \nenter a weight measurement please:\t ; 22 cin >> pound; 23 cout << \t = << (.454 * pound); 24 count ++; cout << endl << Thank you ; 28 return (0); 29 The body of the loop consists of the statements in line 21 to line 24. The loop-counter is initialized in line 18; and the loop-increment statement is in line 24. Exercise CS10* Assuming that all variables are properly defined, find the mistake(s) in the specification of each of the following while structures: a. b. c. while( count < 10 ) while(count < 10) count = 0; while(count < 10 ) cin >> num; count = 0; cout << 2 * num; cin >> num; cin >> num; count = count + 1; cout << 2 * num; cout << 2 * num; count = count + 1; Exercise CS11 Write a C++ code segment to read 50 temperature values in Fahrenheit and to convert them to Celsius. You convert a Fahrenheit temperature to Celsius by using the following formula: Celsius = 5.0 / 9 * (Fahrenheit - 32) Gilbert Ndjatou Page 76

21 Using a Running Total A running total is a variable that is used to compute the sum of the values processed at ach iteration of a loop: it is first initialized to 0 and each new value processed in the loop is added to the previous total. Case Study CS5 Problem Statement Write a program to read 20 integer values and to compute their sum. Program Logic Input: 20 integer values. Output: their sum. Variable: count (int) to count the values: Initial value is 0 (no value is read so far); Final value is 20. value (int) to hold the integer value read. totalvalue (int) to hold the sum of the integer values read so far: Initial value is 0. Algorithm Specification (using a Flowchart) Start totalvalue = 0 count = 0 count < 20 Read value totalvalue = totalvalue + value Write totalvalue count = count + 1 Stop 2013 Gilbert Ndjatou Page 77

22 Figure CS5 Counter-Controlled Iteration using the while Structure Line Number 1 /************************************************************* 2 Program to read twenty integer values and to compute their sum. 3 *************************************************************/ 4 #include <iostream> 5 using namespace std; 6 #define MAXCOUNT int main() 9 10 int count; // to count values 11 value, // to hold the value read 12 totalvalue; // the hold the sum of the values read so far /*------read all values and compute their sum */ 15 totalvalue = 0; 15 count = 0; // no value has been read so far 16 while (count < MAXCOUNT) // repeat twenty times cout << \nenter an integer value please:\t ; 19 cin >> value; 20 totalvalue += value; 21 count++; /*-----print the sum of all the values read */ 25 cout << \n\nthe sum of all the values read is:\t 26 << totalvalue; 27 return (0); 28 Exercise CS12* Each of the following code segments is supposed to read 10 integer values and to compute their sum. What is wrong in each of these code segments? a. b. count = 0; count = 0; while (count < 10 ) while (count < 10 ) sum = 0; cin>>num; cin >> num; sum = sum + num; sum = sum + num; count = count + 1; count = count + 1; cout << endl << sum; cout << endl << sum; 2013 Gilbert Ndjatou Page 78

23 Exercise CS13* Write a C++ code segment to read the test scores of twenty students and to compute and print their average. Exercise CS14 A client has purchased 20 products in a store. Write a C++ code segment to read the unit price and the number of items of each product and to compute and print the total price of all these products. Case Study CS6 Problem Statement Write a program to compute the sum of the first 20 positive even integer values (2, 4,...). Program Logic Input: none. Output: the sum of the first 20 positive integer values. Variable: count (int) to count the values: Initial value is 1; Final value is 21. totalvalue (int) to hold the sum of the integer values processed so far: Initial value is 0. Algorithm Specification (using a Flowchart) Start totalvalue = 0 count = 1 Write totalvalue Stop count < 21 totalvalue = totalvalue + 2 * count count = count Gilbert Ndjatou Page 79

24 Figure CS6 Counter-Controlled Iteration using the while Structure Line Number 1 /************************************************************* 2 Program to read twenty integer values and to compute their sum. 3 *************************************************************/ 4 #include <iostream> 5 using namespace std; 6 #define MAXCOUNT int main() 9 10 int count; // to count the values 11 totalvalue; // the hold the sum of the values so far totalvalue = 0; 14 count = 1; 15 while (count < MAXCOUNT) // repeat twenty times totalvalue += 2 * count; 18 count ++; /*-----print the sum of the first 20 positive even values */ 23 cout << \n\nthe sum of the first 20 positive even values is:\t 24 << totalvalue; 25 return (0); 26 Exercise CS15 Write a C++ code segment to add the first 20 positive multiples of 5 ( ) and to output the result. Using a Running Product and a Counter with the initial value Greater than the final Value A running product is a variable that is used to compute the product of values processed at each iteration of a loop: it is first initialized to 1 and each new value processed in the loop is multiplied to it. Case Study CS7 Problem Statement For a positive integer value n, n factorial (written n!) is the product of all the integer values from 1 to n (1 * 2 * 3 *... * n). Write a program to read a positive integer value n, and to compute and print n! 2013 Gilbert Ndjatou Page 80

25 Program Logic Input: a positive integer value. Output: the product of all positive integer values less than or equal to the value read. Variables: number (int) to hold the value read factorial (int) to hold the product of the integer values: initial value = 1 After Iteration Value of number value of factorial - n 1 1 n x n 2 n x n x (n - 1)... n x n x (n - 1) x... x 2 Algorithm Specification (using a Flowchart) Start Read number factorial = 1 number > 1 Factorial = factorial * number Write factorial number = number - 1 Stop 2013 Gilbert Ndjatou Page 81

26 Figure CS7 Counter-Controlled Iteration using the while Structure Line Number 1 /************************************************************* 2 Program to read a positive integer value and to compute and print its factorial. 3 *************************************************************/ 4 #include <iostream> 5 using namespace std; 6 int main() 7 8 int number, //integer value/loop counter 9 factorial = 1; // to hold the product /* read the integer value */ 12 cout << endl << enter a positive integer value:\t ; 13 cin >> number; /* compute the factorial of the number read */ 16 while (number > 1) factorial *= number; 19 number --; /* print the factorial */ 23 cout << endl << n factorial is:\t 24 << factorial; 25 return (0); 26 The body of the loop consists of the statements in lines 18 and 19. The loop-counter is initialized in line 13; and the loop-increment statement is in line 19. Exercise CS16 Write a C++ code segment to generate the squares (n*n) of the integer values 20, 18, 16, 14,..., 2, starting with 20. Exercise CS17 Write a C++ code segment to read ten integer values and to compute and print their product Gilbert Ndjatou Page 82

27 Logically-Controlled Iteration using the while Structure The solutions of a class of problems require the CPU to repeat the execution of one or more statements as long as a certain condition is true (that means, until a certain condition becomes false). Case Study CS8 Problem Statement A can of soda costs $ 1.00 in a soda machine, and a user buys a can of soda by depositing coins (nickels, dimes or quarters) in the machine until the total value of the coins deposited in the machine is greater than or equal to $ Write a program to read the value of each coin deposited in the soda machine (5, 10, or 25) until the total value of the coins deposited in the machine is greater than or equal to 100. The program will then output the message Thank you for using the soda machine and the amount of the change. Program Logic input: the value of a coin (5, 10 or 25). Output: Note: the message Thank you for using this soda machine and the amount of change. To read the value of each coin deposited in the machine and to compute their sum, we need the following: A variable coinvalue to hold the input value of a coin. A variable totalvalue to hold the sum of the values of the coins read so far. It is initialized to 0 before the body-of-the-loop. The loop condition is: totalvalue < 100. Variables: coinvalue ( int ) totalvalue ( int ) to hold the value of a coin. to hold the total value of the coins deposited so far Gilbert Ndjatou Page 83

28 Algorithm Specification (using a Flowchart) Start totalvalue = 0 totalvalue < 100 Read coinvalue totalvalue = totalvalue + coinvalue Write Thank you Write totalvalue Stop 2013 Gilbert Ndjatou Page 84

29 Figure CS8 Line Number Logically-Controlled Iteration using the while Structure 1 /***************************************************************** 2 Program to monitor the purchase of a can of soda in a soda machine 3 *****************************************************************/ 4 #include <iostream> 5 using namespace std; 6 int main() 7 8 const int SODAPRICE = 100; // the price of a can of soda 9 int coinvalue, // the value of a coin 10 totalvalue = 0; // the total value of the coins 11 while( totalvalue < SODAPRICE ) /* read the value of the next coin */ 14 cout << endl << enter the value of the next coin:\t ; 15 cin >> coinvalue; /* compute the total value of the coins deposited so far */ 18 totalvalue = totalvalue + coinvalue; /*--- say thank you and give the change back to the user */ 22 cout << \n\nthank you for using this soda machine 23 << \n\nyour change is:\t << (totalvalue - SODAPRICE); 24 return ( 0 ); 25 Exercise CS18* The manager of a store is required to make a deposit of $ 10,000 whenever the sum of the daily sales in that store is $ 10,000 or more. Write a C++ code segment to read the daily sales in that store, and to compute their sum until it becomes greater than or equal to $ 10,000. The sum of the daily sales and the message it is time to make a deposit are then printed. Exercise CS19 A store has 120 bags of coffee and every day, the amount of bags sold is deducted from the current stock until it becomes less than or equal to 30. The manager of the store must then place a new order. Write a C++ code segment to read the number of bags sold daily and to update the current stock until it becomes less than or equal to 30. The message place a new order and the amount of bags to purchase in order to have 120 bags are then printed Gilbert Ndjatou Page 85

30 while Structure with one Statement in the Body-of- the-loop When the body-of-the-loop consists of just one statement, the braces may be omitted from the while structure. Case Study CS9 Problem Statement Write a program to compute the smallest power of 5 greater than (1, 5, 25, 125,... ). Program Logic Input: none. Output: the smallest power of 5 greater than Note: To compute the smallest power of 5 greater than 10000, we need: A variable power to hold the powers of 5. It is initialized to 1 before the body-of-the-loop. The current value of variable power is multiplied by 5 in the body-of-the-loop until it becomes greater than The loop condition is therefore: power <= After Value of Iteration variable power x 5 = x 5 = x 5 = Variable: power (int) to hold the powers of Gilbert Ndjatou Page 86

31 Algorithm Specification (using a Flowchart) Start power = 1 power <= Power = power * 5 Stop Figure CS9 Line Number while Structure with a single statement in the body-of-the-loop 1 /*************************************************************** 2 Program to compute the smallest power of 5 greater than ****************************************************************/ 4 #include <iostream> 5 using namespace std; 6 int main() 7 8 int power; // to hold the current power of 5 9 power = 1; // first power of 5 10 while ( power <= 10000) 11 Power = power * 5; /* print the result */ 14 cout << endl << power 15 << is the smallest power of 5 greater than 10000"; 16 return (0); 17 Although the body-of-the-loop consists of just one statement, it could also be specified as a compound statement as follows: while (power <= 10000) power = power * 5; 2013 Gilbert Ndjatou Page 87

32 Sentinel-Controlled Iteration When you want to process a list of values and you do not know how many values are in the list, you can use a sentinel value (dummy value, flag value, or signal value) to mark the end of the list. A sentinel value is in general chosen in such a way that they can not possibly be one of the data items in the list. Example: For a program to read and process a list of test scores received by students in an exam, a sentinel value might be -1 or -99. With -99, the list could be: The data items in the list are accessed and processed one at a time until the sentinel value is accessed. However, the sentinel value must not be processed. The algorithm to perform this type of repetition is specified using a flowchart as follows: Read/Access the first data item data item!= sentinel Process a data item Next-Statement Read/Access the next data item 2013 Gilbert Ndjatou Page 88

33 It is specified using the while structure as follows: <statement-to read/access-the first-value> While(<value-read/access!= sentinel-value>) <processing-statements> <statement-to-read/access-the-next-value> <processing-statements> consists of one or more statements used to process a data item in the list. Case Study CS10 Problem Statement Write a program to read one or more weight measurements in pounds and to convert them into kilograms. The dummy value is entered to end the input of the weight measurements. Note that 1 Lb =.454 Kgs. Program Logic Input: one or more weight measurement (in Lbs) followed by the sentinel value Output: weight measurements in (Kgs). Variables: pound (double) to hold a weight measurement in Lbs Gilbert Ndjatou Page 89

34 Algorithm Specification (using a Flowchart) Start Read pound pound!= -99 Write.454 * pound Read pound Write Thank you Stop Figure CS10 Sentinel-Controlled Iteration using the while Structure Line Number 1 /************************************************************* 2 Program to read one or more weight measurements in pounds and 3 to convert them into kilograms. 4 *************************************************************/ 5 #include <iostream> 6 #include <iomanip> 7 using namespace std; 8 #define COEFICIENT #define DUMMYWEIGHT int main() double pound, // the current weight measurement in Lbs cout << fixed << showpoint) << setprecision(2); Gilbert Ndjatou Page 90

35 17 /*read the first weight measurement or the dummy value -99.0*/ 18 cout << endl << Enter a weight measurement or to terminate:\t ; 19 cin >> pound; 20 while (pound!= DUMMYWEIGHT) cout << \t = << (.454 * pound); 23 cout << endl << Enter a weight measurement or to terminate:\t ; 24 cin >> pound; cout << endl << Thank you ; 27 return (0); 28 Exercise CS20* The following program that should read a list of integer values and compute their sum until the sentinel value -99 is read, contains some error(s). Rewrite it with these errors corrected. #include <iostream> using namespace std; #definedummyvalue -99 int main( ) int total, value; cout << \nenter all values to add followed by << DUMMYVALUE << to stop\n ; while ( value!= DUMMYVALUE ) cin >> value; total = total + value; cout << endl << The total of these value is:\t <<total; return( 0 ); Exercise CS21* Write a C++ code segment to read one or more temperature values in Fahrenheit and to convert them to Celsius. The sentinel value -99 marks the end of the input temperatures. You convert a Fahrenheit temperature to Celsius by using the following formula: Celsius = 5.0 / 9 *(Fahrenheit - 32). Exercise CS22 Write a C++ code segment to read the pay rate and the number of hours of work of one or more employees and to compute each employee s gross pay: pay rate * number of hours. The dummy pay rate marks the end of the input values Gilbert Ndjatou Page 91

36 Exercise CS23 Write a C++ code segment to read the test scores of one or more students (until the sentinel value -99 is read) and to compute their average. One-Way Selection using the if Structure A one-way selection structure is used to specify that the execution of one or more actions be performed only if a test performed on a condition is true. It is specified using a flowchart as follows: condition C-Statement Next-Statement C-Statement is one or more statements that specify the operations to be performed when condition evaluates to true. It is specified in C++ language using the if structure as follows: if (<condition>) <Statement> <Next-Statement> where <Statement> is a single statement or a compound statement. Note that when <condition> evaluates to false, no action is performed. Instead, the execution of the program continues with the execution of <Next-Statement> Gilbert Ndjatou Page 92

37 Case Study CS11 Problem Statement A rebate of 5% is given in a store only if the price of the merchandise is at least $ Write a program to read the unit price and the number of quantity of the merchandise, and to compute the amount of the purchase after the rebate if applicable. Program Logic Input: Output: the unit price and the quantity of the merchandise. the total amount of purchase. Variables: unitprice ( double ) to hold the unit price. quantity ( int ) to hold the quantity. purchase ( double ) to hold the amount of purchase. Algorithm Specification (using a Flowchart) Start Read unitprice, quantity purchase = unitprice * quantity purchase >= purchase = purchase -.05 * purchase Write purchase Stop 2013 Gilbert Ndjatou Page 93

38 Figure CS11 Line Number One-Way Selection using the if Structure 1 /**************************************************************** 2 Program to compute the total amount of purchase in a store 3 ****************************************************************/ 4 #include <iostream> 5 using namespace std; 6 int main() 7 8 const double MINPURCHASE = ; //minimum purchase 9 const double RRATE = 0.05; // rebate rate 10 double unitprice, // unit price 11 Purchase // amount of purchase 12 int quantity; // number of quantity /* read in the unit price and the quantity */ 15 cout << \nenter the merchandise unit price:\t ; 16 cin >> unitprice; 17 cout << \nenter its quantity:\t ; 18 cin >> quantity; /* compute the amount of purchase */ 21 purchase = unitprice * quantity; 22 if(purchase >= MINPURCHASE) 23 purchase = purchase - RRATE * purchase; /* print the amount of purchase */ 26 cout << \nyour total amount of purchase is:\t 27 << purchase; 28 return( 0 ); 29 The Null Statement The null statement is a statement that consists of just the semicolon. It says to do nothing, but to continue with the next operation. You may also use the if-else structure and the null statement to specify a one-way selection as follows: if (<condition>) else; <Statement> <Next-Statement> 2013 Gilbert Ndjatou Page 94

39 Example The if structure in lines 22 and 23 of the program in Figure 3.19 could be written as follows: if (purchase >= ) else; purchase = purchase - RRATE * purchase; /* print the amount of purchase */ cout << \nyour total amount of purchase is:\t << purchase; Exercise CS24* What is the output of the following program segment for each of the following input values? a. input value: 4 b. input value: 20 cin >> num; if( num >= 5) pnum = num + 7; cout << endl << pnum=\t << pnum; cout << result=\t << (num - 1); Exercise CS25* greatt, smallt, and somet are integer variables with initial values. Write a C++ code segment to read an integer value and to do the following: add 1 to smallt if the value read is less than 5; then add 2 to greatt if the value read is greater than 20, otherwise subtract 3 from smallt. Then, regardless of the value read add 7 to somet. Exercise CS26 In a department store, a 10% rebate is given for every purchase of at least $ Write a C++ code segment to read the unit price of a product, the quantity purchased, and to compute the amount of the purchase after the rebate if applicable. Exercise CS27 sum, large_ct, middle_ct, and small_ct are integer variables with initial values. Write a C++ code segment to carry out the following tasks: read a value into the integer variable num. if the value read is greater than 100, add 1 to variable large_ct; if it is less than 50, add 2 to variable small_ct; otherwise, add 3 to variable middle_ct; regardless of the value read, add it to variable sum Gilbert Ndjatou Page 95

40 Multiple-Way Selection using if-else Structures A multiple-way selection structure is used to specify that the action to be performed be chosen from a list of three or more actions based on the true values of two or more conditions. A multiple-way selection with four actions and three conditions is specified as follows: condition1 Action-1 condition2 Action-2 condition3 Action-3 Action-4 Next-Action Condition1 is first tested, and if it is true, Action-1 is executed, followed by Next-Action; Otherwise, condition2 is tested, and if it is true, Action-2 is executed, followed by Next-Action; Otherwise, condition3 is tested, and if its true, Action-3 is executed, followed by Next-Action; Otherwise, Action-4 is executed, followed by Next-Action. Note that only one of the actions, Action-1, Action-2, Action-3, or Action-4 is executed. In some situations, Action-4 may also be the null action Gilbert Ndjatou Page 96

41 This behavior is specified in C/C++ using if-else structures as follows: if ( <condition1> ) <Statement-1> else if ( <condition2> ) <Statement-2> else if ( <condition3> ) <Statement-3> else <Statement-4> <Next-Statement> <Statement-1>, <Statement-2>, <Statement-3>, and <Statement-4> are single or compound statements that specify respectively, Action-1, Action-2, Action-3, and Action-4. <Next-Statement> is a statement that specifies the Next-Action. In the case that <Statement-4> is the null statement, this behavior may also be specified as follows: if ( <condition1> ) <Statement-1> else if ( <condition2> ) <Statement-2> else if ( <condition3> ) <Statement-3> <Next-Statement> Case Study CS12 Problem Statement Four letters M, S, D, and W are used to encode the marital status of an individual as follows: M (married), S (single), D (divorced), and W (widowed). Any other character is considered invalid. Write a program to read a marital code and to output the corresponding marital status. Program Logic Input: Output: Variables: mcode (char ) a marital code (capital M, S, D, or W). a marital status. Any other character is considered invalid. to hold the marital code Gilbert Ndjatou Page 97

42 Algorithm Specification (using a Flowchart) Start Read mcode mcode = M Write married mcode = S mcode = D Write single Write divorced Mcode = W Write Widowed Write invalid Stop Figure CS12 Multiple-Way Selection using if-else Structures Line Number 1 /**************************************************************** 2 Program to read a marital code and to output the corresponding 3 marital status. 4 ****************************************************************/ 5 #include <iostream> 6 using namespace std; 2013 Gilbert Ndjatou Page 98

43 7 int main() 8 9 char mcode; // to hold a marital code /* read in a marital code */ 12 cout << \nenter a marital code:\t ; 13 cin >> mcode; /*---determine and print the corresponding marital status --*/ 16 if (mcode == M ) 17 cout << \nthis individual is married ; 18 else if (mcode == S ) 19 cout << \nthis individual is single ; 20 else if (mcode == D ) 21 cout << \nthis individual is divorced ; 22 else if (mcode == W ) 23 cout << \nthis individual is a widow ; 24 else 25 cout << \nthe marital code is invalid ; 26 return( 0 ); 27 Exercise CS28* A florist who sells four different types of bouquets of flowers has numbered them 101, 202, 303, and 404 with their unit prices specified as follows: ITEM NUMBER UNIT PRICE 101 $ $ $ $ 6.50 Write a C++ code segment to read an item number and the number of bouquets of that type purchased, and to compute and print the price of those bouquets. Exercise CS29 A game consists of selecting a positive integer value and rewarding bonuses to the player according to the remainder in the division of that number by 5 as follows: REMAINDER BONUS Write a C++ code segment to read a positive integer value and to output the bonus received by the user Gilbert Ndjatou Page 99

44 Case Study CS13 Problem Statement Write a program to read a test score and to determine and print the corresponding letter grade as follows: A ( score >= 90.00) B (80.00 <= score < 90.00) C (70.00 <= score <80.00) D (60.00 <= score < 70.00), F (score < 60.00). It is assumed the input value is a valid test score. Program Logic Input: a test score (floating-point value in the range 0.00 to ). Output: the corresponding letter grade. Variables: score (double ) to hold a test score. Algorithm Specification Using a Flowchart Start Read score score >= Write A Write B score >= Score >= Write C score >= Write D Write F Stop 2013 Gilbert Ndjatou Page 100

45 Figure CS13 Line Number Multiple-Way Selection using if-else structures 1 /**************************************************************** 2 Program to read a test score and to determine and output the 3 corresponding letter grade. 4 ****************************************************************/ 5 #include <iostream.h> 6 using namespace std; 7 int main() 8 9 double score; // to hold a test score /* read in a test score */ 12 cout << \nenter a test score:\t ; 13 cin >> score; /*----determine and print the corresponding letter grade ---*/ 16 if ( score >= 90.00) 17 cout << \nthe letter grade is:\t << A ; 18 else if (score >= 80.00) 19 cout << \nthe letter grade is:\t << B ; 20 else if (score >= 70.00) 21 cout << \nthe letter grade is:\t << C ; 22 else if (score >= 60.00) 23 cout << \nthe letter grade is:\t << D ; 24 else 25 cout << \nthe letter grade is:\t << F ; 26 return ( 0 ); 26 The multiple-way selection specified in line 16 to line 25 could also be written as follows: if ( score < 60.00) cout << \nthe corresponding letter grade is:\t << F ; else if (score < 70.00) cout << \nthe corresponding letter grade is:\t << D ; else if (score < 80.00) cout << \nthe corresponding letter grade is:\t << C ; else if (score < 90.00) cout << \nthe corresponding letter grade is:\t << B ; else cout << \nthe corresponding letter grade is:\t << A ; 2013 Gilbert Ndjatou Page 101

46 Exercise CS30* What is the output of the following program segment for each of the following input values? a. input value: 3 b. input value: 7 c. input value: 12 cin >> num; if (num < 5) cout << "Red\n"; else if (num < 10) cout << "Yellow\n"; else cout << "Green\n"; cout << blue Exercise CS31* The decision table below shows fines imposed for speeding violations. Write a C++ code segment to read the speed of a car and to output the corresponding fine. Speed <= 50 mph Fine = $ < Speed <= 70 Fine = $ < Speed <= 80 Fine = $ Speed > 80 Fine = $ Exercise CS32 A department store offers rebates to customers according to the price of the items purchased as follows: Price >= $ Rebate = $ $ 50 <= Price < $ Rebate = $ 5.00 $ 20 <= price < $ Rebate = $ 2.00 price < $ No rebate A sale tax of 8.25% is also imposed on the price of each purchase (after the rebate, if any). Write a C++ code segment to read the price of a product and to output the rebate received by the customer, the price (including the rebate), and the amount due (including taxes) Gilbert Ndjatou Page 102

47 Iterations using the for Structure The for structure has the following general syntax: for(<init-expression-list> ; <loop-condition> ; <increment-expression-list>) <body-of-the-loop> It is executed in the same way that the following while structure is executed: <init-expression-statements> while ( <loop-condition> ) <body-of-the-loop> <increment-expression-statements> Example 1. The following counter-controlled iteration (specified using the while structure) may be specified using the for structure in any of the following ways in (a), (b), and ( c ): totalscore = 0; testcount = 0; // no test score has been read so far while (testcount < 20) // repeat twenty times cout << \nenter a test score please:\t ; cin >> score; totalscore + = score; testcount ++; 2013 Gilbert Ndjatou Page 103

48 ( a ) The <init-expression-list> consists of the expression, testcount = 0, The <loop-condition> is testcount < 20, and The <increment-expression-list> consists of the expression, testcount ++. totalscore = 0; for (tescount = 0 ; testcount < 20 ; testcount ++) cout << \nenter a test score please:\t ; cin >> score; totalscore + = score; // repeat twenty times ( b ) Any expression that is evaluated before the <loop condition> is tested for the first time may also be moved into the <init-expression-list>. So, we may also specify the expression totalscore = 0 in the <init-expression-list> as follows: for (totalscore = 0, tescount = 0 ; testcount < 20 ; testcount ++) // repeat twenty times cout << \nenter a test score please:\t ; cin >> score; totalscore + = score; ( c ) Expressions that are evaluated in the <body-of-the-loop> may be moved into the <incrementexpression-list>. However, keep in mind that expressions in an expression list are evaluated from left to right. So, we may move the expression totalscore + = score from the <body-of-the-loop> to the <increment-expression-list> before the loop increment expression as follows: 2013 Gilbert Ndjatou Page 104